What is the derivative of y=x(1+x^2)^{10} ?

The derivative of y=x(1+x^2)^{10} is (1+x^2)^{10}+20x^2(1+x^2)^9

What is the first derivative of y=x(1+x^2)^{10} ?

The first derivative of y=x(1+x^2)^{10} is (1+x^2)^{10}+20x^2(1+x^2)^9

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derivative of

y = x (1 + x^{2})^{10}

(1 + x^{2})^{10} + 20 x^{2} (1 + x^{2})^{9}

Solution steps

\frac{d}{d x} (x (1 + x^{2})^{10})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = (1 + x^{2})^{10}

= \frac{d x}{d x} (1 + x^{2})^{10} + \frac{d}{d x} ((1 + x^{2})^{10}) x

\frac{d x}{d x} = 1

\frac{d}{d x} ((1 + x^{2})^{10}) = 20 x (1 + x^{2})^{9}

= 1 \cdot (1 + x^{2})^{10} + 20 x (1 + x^{2})^{9} x

Simplify

1 \cdot (1 + x^{2})^{10} + 20 x (1 + x^{2})^{9} x : (1 + x^{2})^{10} + 20 x^{2} (1 + x^{2})^{9}

= (1 + x^{2})^{10} + 20 x^{2} (1 + x^{2})^{9}

x \to 0 - lim (\frac{x}{x})

\int (3 y - 4 y^{3})^{3} d y

y = ln (\frac{x}{2}), at x = 4

\int \frac{t ^{2}}{t + 1} d t

\frac{d}{d x} (l x)

What is the derivative of y=x(1+x^2)^{10} ?
The derivative of y=x(1+x^2)^{10} is (1+x^2)^{10}+20x^2(1+x^2)^9
What is the first derivative of y=x(1+x^2)^{10} ?
The first derivative of y=x(1+x^2)^{10} is (1+x^2)^{10}+20x^2(1+x^2)^9